Let $a(x)=-12x^5-2x^3-9x$, and $b(x)=3x^4+x^2+1$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Solution: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{-12x^5-2x^3-9x}{3x^4+x^2+1}$ : We divide ${3x^4}$ into ${-12x^5}$ to get ${-4x}$ : $ \hphantom{1567|1444477} {{-4x}}\\ {{{3x^4}+x^2+1}}|\overline{{-12x^5}-2x^3-9x}\\ \hphantom{37...88........|}\llap{-}\underline{(-12x^5-4x^3-4x)}\\ \hphantom{37|3....99888889......}{+2x^3-5x }\\ $ [What did we do here?] The process stops here because $3x^4+x^2+1$ is a polynomial of the fourth degree and $2x^3-5x$ is a polynomial of the third degree. So it follows that ${r(x)}={2x^3-5x}$, ${q(x)}={-4x}$, and $ \dfrac{-12x^5-2x^3-9x}{3x^4+x^2+1}={-4x}+\dfrac{{2x^3-5x}}{3x^4+x^2+1}$ To conclude, $q(x)=-4x$ $r(x)=2x^3-5x$